G11 Chemistry - Atomic Structure & Periodic Properties PDF

 1 ATOMIC STRUCTURE AND PERIODIC PROPERTIES OF THE ELEMENTS Unit Outcomes At the end of this unit, you will be able to: ) discuss the historical development of atomic structure ) explain the experimental observations and inferences made by some famous sci...

 1 ATOMIC STRUCTURE AND PERIODIC PROPERTIES OF THE ELEMENTS Unit Outcomes At the end of this unit, you will be able to: ) discuss the historical development of atomic structure ) explain the experimental observations and inferences made by some famous scientists to characterize the atom ) list the subatomic particles ) identify atomic mass and isotope terms ) explain electromagnetic radiation, atomic spectra and Bohr models of the atom ) compute the calculations involving atomic structure ) describe the quantum mechanical model of the atom and the related postulates and principles ) demonstrate periodic law and how electronic configurations of atoms are related to the orbital diagrams and can explain periodic trends ) describe scientific enquiry skills along this unit: inferring, predicting, classifying, comparing and contrasting, communicating, asking questions and making generalizations. UNIT 1 1  CHEMISTRY GRADE 11 Start-up Activity Discuss the following questions in groups, and present your ideas to the whole class. 1. What are the basic building blocks of the following substances? a. water b. chalk c. sugar d. table salt Are the basic building blocks of these substances the same or different? 2. Why do different materials show different properties? For example, materials such as woods can burn; an iron nail can rust; table salt dissolves in water, etc. 1.1  At the end of this section, you will be able to: ) compare the views of different philosophers on the nature of matter. In Grade 9, you learned about the historical development of the atomic theories of matter. In this unit, we will briefly revise it and discuss in detail about early experiments to characterize an atom, the atomic spectra, different models of an atom, etc. Activity 1.1 Form a group and discuss the following questions, then present your views to the whole class. 1. How do the beliefs about the structure of matter evolve? 2. Describe the early developments leading to the modern concept of the atom. 3. Can we see atoms with our naked eyes? The philosophers of ancient Greece wondered about the composition of matters: is matter continuously divisible into ever smaller and smaller pieces, or is there an ultimate limit? Although most philosophers, including Plato and Aristotle, believed that matter is continuous, Democritus disagreed. The Greek philosopher Democritus (460–370 BC) suggested that if you divided matter into smaller and smaller pieces, you would eventually end up with tiny, indestructible particles called atomos, or “atoms”, meaning “indivisible”. 2 UNIT 1  Dalton’s Atomic Theory and the Modern Atomic Theory His ideas were based on philosophical speculation rather than experimental evidence. The ideas of Democritus were not widely accepted until 1808. After nearly 2000 years, John Dalton, developed an atomic theory that had gained broad acceptance. 1.2     At the end of this section, you will be able to: ) state postulates of Dalton’s atomic theory ) state postulates of the modern atomic theory ) state the laws of conservation of mass, definite proportions, multiple proportions and the basis of each of these laws ) use postulates of Dalton’s atomic theory to explain the laws of definite and multiple proportions ) evaluate postulates of Dalton’s and the modern atomic theories.        Activity 1.2 Recall your Grade 9 knowledge in order to answer the following questions in groups, and share your responses with the whole class. 1. Describe the five postulates of Dalton’s atomic theory. 2. How do the scientific ideas develop based on previous scientific findings? Scientific laws usually develop based on previous scientific findings. The laws that are the basis for Dalton’s atomic theory are the law of conservation of mass and the law of definite proportions. What is the law that Dalton formulated based on the law of conservation of mass and the law of definite proportions? Write its statement in your notebook. UNIT 1 3  CHEMISTRY GRADE 11 Activity 1.3 In your Grade 9 chemistry lesson, you learned about the law of conservation of mass and the law of definite proportion. Form a group and discuss the following questions, then present your responses to the whole class. 1. The mass of a piece of wood before and after it is burnt to ashes is not the same. Does this show that mass is created or destroyed? 2. What would be the mass of products if the burning of wood was carried out in a closed container? 3. Sugar consists of C, H, and O atoms. When a certain amount of sugar is burned in a crucible, it changes from white sugar to black carbon. Where has the hydrogen and oxygen gone? Example 1.1 1. What mass of hydrogen and oxygen can be obtained from a. 18.0 g of water b. 1.00 g of water Solutions: Water is always 11.2% hydrogen and 88. 8% oxygen by mass, so: a. mass of hydrogen in 18.0 g water = 0.112 × 18.0 g = 2.02 g mass of oxygen in 18.0 g water = 0.888 × 18.0 g = 15.98 g; or = 18.0 g – 2.02 g = 15.98 g b. mass of hydrogen in 1.00 g water = 0.112 × 1. 00 g = 0.112 g mass of oxygen in 1.0 g water = 0.888 × 1.0 g = 0.888 g; or = 1.0 g – 0.112 g = 0.888 g 2. The following data were collected for several compounds of nitrogen and oxygen: Compound A Compound B Compound C Mass of nitrogen that combines with 1 g of oxygen 1.750 g 0.8750 g 0.4375 g Show how these data illustrate the law of multiple proportions. 4 UNIT 1  Dalton’s Atomic Theory and the Modern Atomic Theory Solution: For the law of multiple proportions to hold, the ratios of the masses of nitrogen combining with 1 gram of oxygen in each pair of compounds should be small whole numbers. We therefore compute the ratios as follows: A 1.750 4 B 0.875 2 C 0.4375 1 = = = = = = C 0.4375 1 C 0.4375 1 C 0.4375 1 These results support the law of multiple proportions. Exercise 1.1 1. List the postulates of Dalton’s that continue to have significance (are retained in the modern atomic theory). 2. How does the atomic theory account for the fact that when 1.00 g of water is decomposed into its elements, 0.112 g of hydrogen and 0.888 g of oxygen are obtained regardless of the source of the water?     Activity 1.4 Remember what you have learned in Grade 9 and discuss the following questions in a group of three or four, then present your responses to the whole class. 1. Which of Dalton’s postulates about atoms are inconsistent with later observations? Do these inconsistencies mean that Dalton was wrong? 2. Is Dalton’s model still useful? Most of the experiments conducted during the development of the modern atomic theory will be discussed in Sections 1.3-1.6. In this section, generalizations derived from the experiments are presented as postulates of the modern atomic theory. The modern atomic theory is generally said to begin with John Dalton. Dalton’s work was mainly about the chemistry of atoms how they combine to form new compounds rather than about the internal structure of atoms. The modern theories about the physical structure of atoms did not begin until J.J. Thomson discovered the electron in 1897. UNIT 1 5  CHEMISTRY GRADE 11 Exercise 1.2 1. Describe the limitations of Dalton’s atomic theory. 2. Explain the postulates of the modern atomic theory. 3. List the three fundamental laws of chemistry. 4. How does the modern atomic theory explain the three fundamental laws of chemistry? 1.3    At the end of this section, you will be able to: ) discuss the discovery of the electron ) describe the properties of cathode rays ) define the terms: radioactivity, radioactive decay and radio-isotope ) describe the common types of radioactive emissions ) illustrate the alpha scattering experiment, and summarize and interpret the major contribution of experiments of Thomson, Millikan and Rutherford concerning atomic structure.  Activity 1.5 Think back to Grade 9 and reflect on the following questions in a group of four. Then, present your responses to the whole class. a. How were electrons discovered? b. Are cathode rays visible to the naked eye? 6 UNIT 1  Early Experiments to Characterize the Atom Historical note: J.J. Thomson (1856-1940) was a British physicist and Nobel laureate. Sir Joseph John Thomson was born near Manchester, England, and educated at Owens College (now part of Victoria University of Manchester) and Trinity College, University of Cambridge. At Cambridge he taught mathematics and physics, served as Cavendish Professor of Experimental Sir Joseph Thomson Physics, and was (1918-40) master of Trinity College. Thomson’s work on cathode rays led to the discovery of the electron in 1897. His later work with positive ion beams led to a method of separating atoms and molecules by mass and the discovery of neon. Thomson was awarded the Nobel Prize in Physics in 1906 and was knighted in 1908. Cathode Rays One of the first experiments on subatomic particles was carried out by the English physicist J.J. Thomson in 1897. Figure 1.1 shows an experimental apparatus similar to the one used by Thomson. In this apparatus, two electrodes from a high-voltage source are sealed into a glass tube from which the air has been evacuated. The negative electrode is called the cathode; the positive one, the anode. When the high-voltage current is turned on, the glass tube emits a greenish light. Experiments showed that this greenish light is caused by the interaction of the glass with cathode rays, which are rays that originate from the cathode. Figure 1.1: Formation of cathode rays UNIT 1 7  CHEMISTRY GRADE 11 After the cathode rays leave the negative electrode, they move toward the anode, where some rays pass through a hole to form a beam (Figure 1.1). This beam bends away from the negatively charged plate and toward the positively charged plate. What did Thomson conclude from this observation? Figure 1.2 shows a similar experiment, in which cathode rays are seen to bend when a magnet is brought toward them. Thomson showed that the characteristics of cathode rays are independent of the material making up the cathode. From such evidence, he concluded that a cathode ray consists of a beam of negatively charged particles (or electrons) and that electrons are constituents of all matter. Figure 1.2: Bending cathode rays using a magnet By measuring the amount of deflection of a cathode ray beam in electric and magnetic fields of known strengths, Thomson was able to calculate the ratio of the mass of an electron, me, to its charge, e. The number he came up with is – 5.686 × 10–12 kg C−1 (kilograms per coulomb). However, he could not obtain either the mass or the charge separately. In 1909, Robert A. Millikan, an American physicist, measured the charge of the electron by measuring the effect of an electrical field on the rate at which charged oil drops fell under the influence of gravity. Based on careful experiments, Millikan established the charge on an electron as e = –1.602 × 10–19 C. He used this value and Thomson’s mass/charge ratio to calculate an electron’s mass to be 9.109 × 10–31 kg. m me = e × e =−5.686 × 10−12 kg C−1 × − 1.602 × 10−19 C e = 9.109 × 10−31 kg 8 UNIT 1  Early Experiments to Characterize the Atom        Activity 1.6 Answer the following questions individually, then share your answers with the whole class. 1. Describe the properties of cathode rays. 2. Which rays are used to see whether bones are broken or not? Radioactivity Radioactivity or radioactive decay is the spontaneous emission of particles and/or radiation from the unstable nuclei of certain atoms such as uranium, radium, etc. Does radioactivity support Dalton’s idea of atoms? Shortly after the discovery of radioactivity, three types of rays were identified in the emissions from radioactive substances. Two are deflected by oppositely charged metal plates (Figure 1.3). Alpha (α) rays consist of positively charged particles, called α particles. They have a mass of about four times that of a hydrogen atom and a charge twice the magnitude of an electron; they are identical to helium nuclei. Beta (β) rays, or β particles, are electrons coming from inside the nucleus and are deflected by the negatively charged plate. The third type of radioactive radiation consists of high-energy rays called gamma (γ) rays. They have no charge and are not affected by an external electric or magnetic field. Figure 1.3: Three types of ray emitted by radioactive elements UNIT 1 9  CHEMISTRY GRADE 11 The Discovery of the Nucleus Thomson proposed a “plum-pudding” model (Figure 1.4) for the atom in which the electrons and protons were randomly distributed in a positively charged cloud like plums in a pudding. Figure 1.4: Thomson’s “plum-pudding” model of an atom In1911, Ernest Rutherford worked with Thomson to test this model. In Rutherford’s experiment, positively charged particles were aimed at a thin sheet of gold foil (Figure 1.5). If the Thomson model were correct, the particles would travel in straight paths through the gold foil. Rutherford was greatly surprised to find that some of the particles were deflected as they passed through the gold foil, and a few particles were deflected so much that they went back in the opposite direction. Figure 1.5: (a) α-particles aimed at a piece of gold foil, (b) Magnified view of α-particles passing through and being deflected by nuclei 10 UNIT 1  Early Experiments to Characterize the Atom Activity 1.7 Form a group and discuss Rutherford’s experiment as shown in Figure 1.5 (a) and (b) and answer the following questions: 1. Why did most of the α-particles pass through the foil undeflected? 2. Why did only a small fraction of the α-particles show a slight deflection? 3. Why didn't all α-particles bounce at an angle of 180°? 4. Based on the findings of Rutherford’s experiment, what is your conclusion about an atom? 5. When Rutherford’s co-workers bombarded gold foil with α-particles, they obtained results that overturned the existing (Thomson) model of the atom. Explain. Report your responses to the whole class.    The neutron was also discovered by alpha-particle scattering experiments. When beryllium metal is irradiated with alpha rays, a strongly penetrating radiation is obtained from the metal. In 1932 the British physicist James Chadwick (1891– 1974) showed that this penetrating radiation consists of neutral particles, called neutrons. The neutron is a nuclear particle having a mass almost identical to that of the proton but with no electric charge. The mass of a neutron, mn = 1.67493 × 10–27 kg, which is about 1840 times the mass of an electron. Activity 1.8 Discuss the following questions in groups, and present your responses to the whole class. 1. A sample of a radioactive element is found to be losing mass gradually. Explain what is happening to the sample. 2. Describe the experimental basis for believing that the nucleus occupies a very small fraction of the volume of the atom. UNIT 1 11  CHEMISTRY GRADE 11 1.4  At the end of this section, you will be able to: ) describe the make-up of the nucleus ) define atomic mass ) define isotope, and calculate the relative atomic mass (atomic mass) of naturally occurring isotopic elements.       In 1919, Rutherford discovered that hydrogen nuclei, or what we now call protons, form when alpha particles strike some of the lighter elements, such as nitrogen. A proton is a nuclear particle having a positive charge equal in magnitude to that of the electron. A proton has a mass of mp = 1.67262 × 10–27 kg, which is about 1840 times the mass of electrons. The protons in a nucleus give the nucleus its positive charge. Table 1.1 compares the relative masses and charges of the three subatomic particles (note that "amu" stands for "atomic mass unit", which is equal to 1 the mass of an 12 atom of carbon-12). Table 1.1: Properties of subatomic particles Particle Actual mass (kg) Relative mass Actual charge Relative (amu) (C) charge Proton (p) 1.672622 × 10-27 1.007276 1.602 × 10-19 +1 Neutron (n) 1.674927 × 10 1.008665 -27 0 0 Electron (e−) 9.109383 × 10 5.485799 × 10-4 -1.602 × 0-19 -31 -1 The atomic number (Z) of an element equals the number of protons in the nucleus of each of its atoms. All atoms of a particular element have the same atomic number, and each element has a different atomic number from that of any other element. The total number of protons and neutrons in the nucleus of an atom is its mass number (A). The mass number and atomic number of an element X are often written with the symbol, 12 UNIT 1  Make-up of the Nucleus Activity 1.9 Based on your previous knowledge, write the similarities and differences between the pairs of atomic notations and present your responses to the whole class. a. 29 30 14 Si and 14 Si b. 146 C and 147 N 79 81 79 79 c. 35 Br and 35 Br d. 35 Br − and 36 Kr    All atoms of an element are identical in atomic number but not in mass number. Isotopes of an element are atoms that have different numbers of neutrons and different mass numbers. For example, all carbon atoms have six protons in the nucleus (Z = 6) but only 98.89 % of naturally occurring carbon atoms have six neutrons in the nucleus (A = 12). A small percentage (1.11 %) have seven neutrons in the nucleus (A = 13), and even fewer (less than 0.01 %) have eight (A = 14). Hence, carbon has three naturally occurring isotopes: 12C, l3C, and 14C. Most elements found in nature are mixtures of isotopes. The average mass for the atoms in an element is called the atomic mass of the element and can be obtained as averages over the relative masses of the isotopes of each element, weighted by their observed fractional abundances. If an element consists of n isotopes, of relative masses A1, A2…An and fractional abundances of f1, f2… fn, then the average relative atomic mass (A) of the element is: A = A1 f1 + A2 f2 + … + An fn. UNIT 1 13  CHEMISTRY GRADE 11 Example 1.2 There are two naturally occurring isotopes of silver. Isotope 107Ag (106.90509 amu) accounts for 51.84% of the total abundance, and isotope 109Ag (108.90476) accounts for the remaining 48.16%. Calculate the atomic mass of silver? Solution: Find the portion of the atomic mass from each isotope: Portion of atomic mass from 107Ag: = isotopic mass × fractional abundance = 106.90509 amu × 0.5184 = 55.42 amu Portion of atomic mass from Ag: = 108.90476 amu × 0.4816 = 52.45 amu 109 Find the atomic mass of silver: Atomic mass of Ag = 55.42 amu + 52.45 amu = 107.87 amu Exercise 1.3 1. How many protons and neutrons are in the nucleus of each of the following atoms? a. 27 b. 32 64 207 13 Al 16 S c. 30 Zn d. 82 Pb 2. Element X is toxic to humans in high concentration but essential to life at low concentrations. It has four naturally occurring isotopes that contain 24 protons. Identify element X and give the atomic symbol for the isotopes with 26, 28, 29, and 30 neutrons. 3. Naturally occurring boron consists of two isotopes, 10B and 11B, with the isotopic masses 10.013 amu and 11.009 amu, respectively. The observed atomic mass of boron is 10.811 amu. Calculate the abundance of each isotope 4. The two naturally occurring isotopes of lithium, lithium-6 and lithium-7, have masses of 6.01512 amu and 7.01600 amu, respectively. Which of these two occurs in greater abundance? 14 UNIT 1  Electromagnetic Radiation and Atomic Spectra 1.5  At the end of this section, you will be able to: ) characterize electromagnetic radiation (EMR) in terms of wavelength, frequency and speed ) calculate the wavelength and frequency of EMR ) explain the dual nature of light ) describe emission spectra of atoms as consisting a series of lines ) define a photon as a unit of light energy ) distinguish how the photon theory explains the photoelectric effect ) identify the relationship between a photon absorbed and an electron released ) state Bohr’s assumption of energy of the electron in an hydrogen atom ) calculate the radius of electron orbit, the electron velocity and the energy of an electron using Bohr's model ) explain that the line spectrum of hydrogen atom demonstrates the quantized nature of the energy of its electron ) explain that atoms emit or absorb energy when they undergo transitions from one state to another ) compose the limitations of Bohr’s theory.  Activity 1.10 Form a group and discuss the following questions, then present your responses to the whole class. 1. Explain how energy travels in space. 2. What is the importance of electromagnetic radiation (EMR) in chemistry? 3. What are the common features of different energy sources? UNIT 1 15  CHEMISTRY GRADE 11 In 1873, James Clerk Maxwell proposed that light consists of electromagnetic waves. According to his theory, an electromagnetic wave has an electric field and magnetic field components. These two components vibrate in two mutually perpendicular planes (Figure 1.6). EMR is the emission and transmission of energy in the form of electromagnetic waves. Figure 1.6: The electric field and magnetic field components of an electromagnetic waves Electromagnetic waves have three primary characteristics: wavelength, frequency and speed. Wavelength (λ, Greek lambda), is the distance the wave travels during one cycle (Figure 1.7). It is expressed in meters (m) and often, for very short wavelengths, in nanometers (nm), picometers (pm), or angstrom (Å). Frequency (ν, Greek letter nu) is the number of cycles the wave undergoes per second and is expressed in units of l/second (1/s; also called hertz, Hz). Figure 1.7: Frequency of waves 16 UNIT 1  Electromagnetic Radiation and Atomic Spectra Wave has a speed which depends on the type of wave and the nature of the medium through which it is traveling (for example, air, water, or a vacuum). The speed (c) of a wave is the product of its wavelength and its frequency: c = vλ (1.1) In a vacuum, electromagnetic waves travel at 3 ´ 10 m/s, which is a physical constant 8 called the speed of light. EMR comes in a broad range of frequencies called the electromagnetic spectrum (Figure 1.8). A rainbow is an example of a continuous spectrum. Different wavelengths in visible light have different colors from red (λ = 750 nm) to violet (λ = 380 nm). Radiation provides an important means of energy transfer. For instance, the energy from the Sun reaches the Earth mainly in the form of visible and ultraviolet radiation. The glowing coals of a fireplace transmit heat energy by infrared radiation. In microwave ovens, microwave radiation is used to heat water in foods, causig the food to cook quickly. Figure 1.8: The electromagnetic spectrum UNIT 1 17  CHEMISTRY GRADE 11 Example 1.3 Ethiopian National Radio, Addis Ababa station broadcasts its AM signal at a frequency of 2400 kHz. What is the wavelength of the radio wave expressed in meters? Solution: We obtain the wavelength of the radio wave by rearranging Equation 1.1 c 3.00 × 108 m / s so, λ = = = 125.0 m v 2.4 ×106 s Exercise 1. 4 1. The most intense radiation emitted by the Earth has a wavelength of about 10.0 µm. What is the frequency of this radiation in hertz? 2. Addis Ababa Fana FM radio station, broadcasts electromagnetic radiation at a frequency of 98.1 MHz. What is the wavelength of the radio waves, expressed in meters?    Why are waves treated as particles? In 1900, Max Planck, the German physicist, came to an entirely new view of matter and energy. He made a revolutionary proposal, energy like matter is discontinuous. According to Planck, atoms and molecules could emit or absorb energy only in discrete quantities, like small packages or bundles. Each of these small “packets” of energy is called a quantum. The energy of a quantum is proportional to the frequency of the radiation. The energy E of a single quantum is given by: E = hν (1.2) Where h is called Planck’s constant and ν is the frequency of radiation. The value of Planck’s constant is 6.63 ´ 10-34 J. s. 18 UNIT 1  Electromagnetic Radiation and Atomic Spectra Since ν = c/λ, Equation 1.2 can also be expressed as: hc E= (1.3) λ According to quantum theory, energy is always emitted or absorbed in integral multiples of hν; for example, hν, 2hν, 3hν, etc. A system can transfer energy only in whole quanta. Thus, energy seems to have particulate properties. Example 1.4 The blue color in fireworks is often achieved by heating copper (I) chloride (CuCl) to about 1200 °C. Then the compound emits blue light having a wavelength of 600 nm. What is the increment of energy (the quantum) that is emitted at 600 nm by CuCl? Solution: The quantum of energy can be calculated from the Equation 1.2: E = hv The frequency (ν) for this case can be calculated as follows: c v= λ 3.0 × 108 ms -1 = 6.0 × 10-7 m = 0.5 × 1015 s -1 So, E = hv = 6.63 × 10-34 J. s × 0.50 × 1015s-1 = 3.315 × 10-19 J = 3.32 × 10−19 J The Photoelectric Effect In 1905, Albert Einstein used the quantum theory to explain the photoelectric effect. The photoelectric effect is a phenomenon in which electrons are ejected from the surface of certain metals exposed to light of at least a certain minimum frequency, called the threshold frequency, νo. UNIT 1 19  CHEMISTRY GRADE 11 These observations can be explained by assuming that EMR is quantized (consists of photons), and the threshold frequency represents the minimum energy required to remove the electron from the metal’s surface. Photons are particles of light or energy packet. The minimum energy required to remove an electron is: Eo = hνo (1.4) Where Eo is the minimum energy (of the photon), and νo, the threshold frequency. A photon with energy less than Eo (ν < νo) cannot remove an electron, or a light with a frequency less than the νo produces no electrons. On the other hand, if a light has ν > νo, the energy in excess of that required to remove the electron is given to the electron as kinetic energy (KE): KEe = ½ mv2 = hν - hνo (1.5) Where KEe is the kinetic energy of an electron, m is mass of an electron, v is the velocity of an electron, hν, is the energy of an incident photon, and hνo is the energy required to remove an electron from the metal’s surface. Intensity of light is a measure of the number of photons present in a given part of the beam: a greater intensity means that more photons are available to release electrons (as long as ν > νo for the radiation). In his theory of relativity in 1905, Einstein derived the famous equation: E = mc2 (1.6) Rearranging this equation, we have E m = 2 (1.7) c Where E is energy, m is mass, and c is speed of light. The main significance of Equation 1.7 is that energy has mass. Using this equation, we can calculate the mass associated with a given quantity of energy or the apparent mass of a photon. For electromagnetic radiation of wavelength, λ, the energy of each photon is given by the expression: hc Ephoton = (1.8) λ 20 UNIT 1  Electromagnetic Radiation and Atomic Spectra Then, the apparent mass of a photon of light with wavelength is given by: E m= 2 = ( )hc λ = h (1.9) c c2 cλ We can summarize the important conclusions from the work of Planck and Einstein as follows: Energy is quantized. It can occur only in discrete units called photon or quanta. EMR, which was previously thought to exhibit only wave properties, seems to show certain characteristics of particulate matter as well. This phenomenon is sometimes referred to as the dual nature of light and is illustrated Figure 1.9. Figure 1.9: The dual nature of light Example 1.5 1. Compare the wavelength for an electron (mass = 9.11 × 10-31 kg) traveling at a speed of 1.00 × 107 m/s with that for a ball (mass = 0.10 kg) traveling at 35 m/s. Solution: We use the equation m = h/cλ, Where h = 6.63 × 10-34 J.s = 6.63×10-34 kg. m2.s Since 1 J = 1 kg. m2/s2 For the electron: 6.63 × 10−34 kg.m 2.s λ= 9.11× 10−31 kg × 1.00 × 107 m s = 7.28 × 10−11 m = 7.28 nm For the ball: 6.63 × 10−34 kg.m 2.s λ = 1.89 × 10−34 m = = 189 × 10−25 nm 0.10 kg × 35 m / s UNIT 1 21  CHEMISTRY GRADE 11 2. The maximum kinetic energy of the photoelectrons emitted from a give metal is 1.5 × 10–20 J when a light that has a 750 nm wavelength shines on the surface. Determine the threshold frequency, νo , for this metal. Calculate the corresponding wavelength, λo. Solution: a) Determination of the threshold frequency, νo Solve for v from c = v ´ λ Thus, v = c/λ = (3.0 ´ 108 m/s)/7.5 ´ 10-7 m) = 0.4 ´ 1015 s-1 = 4.0 ´ 1014 s-1 Similarly, rearrange Equation 1.5 and solve vo hv − KE vo = h 6.63 ×10−34 J.s × 4.0 × 1014 s −1 − 1.5 × 10 –20 J = 6.63 ×10−34 J. s 26.52 ×10−20 J − 1.5 × 10 –20 J 24.02 × 10−20 J = = 6.63 ×10−34 J. s 6.63 × 10−34 J.s = 3.77 × 1014 s −1 Therefore, a frequency of 3.77 ´ 1014 Hz is the minimum (threshold) required to cause the photoelectric effect for this metal. b) Calculate the corresponding wavelength λo : c λo = v° 3.0 × 108 m s = 3.77 × 1014 s -1 = 0.796 × 10-6 m = 796 nm 22 UNIT 1  Electromagnetic Radiation and Atomic Spectra Exercise 1.5 1. The following are representative wavelengths in the infrared, ultraviolet, and -ray regions of the electromagnetic spectrum, respectively: 1.0 ´ 10-6 m, 1.0 ´ 10-8 m, and 1.0 ´10-10 m. a. What is the energy of a photon of each radiation? b. Which has the greatest amount of energy per photon? c. Which has the least? 2. A clean metal surface is irradiated with light of three different wavelengths λ1, λ 2, and λ 3. The kinetic energies of the ejected electrons are as follows: λ1: 7.2 ´ 10 -20 J; λ2: approximately zero; λ 3: 5.8 ´ 10 -19 J. Which light has the shortest wavelength and which has the longest wavelength? Determine the threshold frequency, ν o, for this metal. 3. The minimum energy required to cause the photoelectric effect in potassium metal is 3.69 ´ 10 -19 J. Will photoelectrons be produced when visible light, 520 nm and 620 nm, shines on the surface of potassium? What is or are the velocities of the ejected electron/s?   Activity 1.11 Discuss the following ideas in pairs and share with the rest of the class. 1. Why do you observe different colors when you are watching fireworks? 2. As it is shown in the Figure 1.10, when compounds of the alkali metals: lithium, sodium, and potassium are excited in the gas flames they give different colored flames. Why do they emit different colors? Figure 1.10: Flame colors of lithium, sodium, and potassium compounds UNIT 1 23  CHEMISTRY GRADE 11 Atomic or line spectra are produced from the emission of photons of electromagnetic radiation (light). Different kinds of spectrum are observed when an electric discharge, or spark, passes through a gas such as hydrogen. The electric discharge is an electric current that excites, or energizes, the atoms of the gas. More specifically, the electric current transfers energy to the electrons in the atoms raising them to excited states. The atoms then emit the absorbed energy in the form of light as the electrons return to lower energy states. When a narrow beam of this light is passed through a prism, we do not see a continuous spectrum, or rainbow, as sunlight does. Rather, only a few colors are observed, displayed as a series of individual lines. This series of lines is called the element’s atomic spectrum or emission spectrum. The wavelengths of these spectral lines are characteristic of the element producing them, and used for their identification. For example, the emission (line) spectrum of hydrogen atom is show in Figure 1.11. Figure 1.11: The hydrogen line spectrum, containing only a few discrete wavelengths Changes in energy between discrete energy levels in hydrogen will produce only certain wavelengths of emitted light, as shown in Figure 1.12. For example, a given change in energy from a high to a lower level would give a wavelength of light which can be expressed using Planck’s equation: ∆E = hν = hc/λ 24 UNIT 1  Electromagnetic Radiation and Atomic Spectra Figure 1.12: A change between two discrete energy levels emits a photon of light      Activity 1.12 Form a group and discuss the following questions. Share your ideas with the rest of the class. The nature of the nucleus of the atom was explained by Rutherford. However, he was not able to explain the position and velocity of electrons in the atom. 1. Is it possible to know the exact location of an electron in the atom? Defend your suggestion. 2. Explain why an electron does not enter the nucleus, even though they are oppositely charged. In 1913, Niels Bohr (a Danish physicist) explained why the orbiting electron does not radiate energy as it moves around the nucleus. He introduced the fundamental idea that the absorption and emission of light by hydrogen atoms was due to energy changes of the electrons within the atoms. The fact that only certain frequencies are absorbed or emitted by an atom tells us that only certain energy changes are possible. Thus, energy changes in an atom are quantized. UNIT 1 25  CHEMISTRY GRADE 11 Bohr used Planck’s and Einstein’s ideas about quantized energy and proposed the following assumptions: 1. The electron in an hydrogen atom travels around the nucleus in a circular orbit. 2. The energy of the electron in an atom is proportional to its distance from the nucleus. The further an electron is from the nucleus, the more energy it has. 3. Only limited number of orbits with certain energies are allowed. This means, the orbits are quantized. 4. The only orbits that are allowed are those for which the angular momentum of the electron is an integral multiple of h/2π. 5. As long as an electron stays in a given orbital, it neither gains or losses energy. That means, the atom does not change its energy while the electron moves within an orbit. 6. The electron moves to a higher energy orbit only by absorbing energy in the form of light, and emitting light when it falls to a lower energy orbit. The energy (photon) of the light absorbed or emitted is exactly equal to the difference between the energies of the two orbits. Furthermore, Bohr showed that the radii, r, of the permitted orbits or energy levels for an atom of hydrogen atom are related to Planck’s constant, h, the electron’s charge, e, and its mass, m. Consider hydrogen atom with an electron with constant speed, v, circulating the nucleus in an orbit of radius, r. The total energy of the electron is the sum of the kinetic energy (energy of movement) and the potential energy (energy of position): 1 2 e2 E = mv + (1.10) 2 r For an electron to exist in a stable orbit of a constant radius, the centripetal force (attracting the electron to the nucleus), e2/r2, and the centrifugal force (pulling away the electron from the nucleus), mv2/r must be equal: e2 mv 2 = (1.11) r2 r Bohr then introduced an additional requirement that the angular momentum, mvr, of the electron can take only certain permitted values, that is, an integral multiple of h/2π. This requirement is called a quantum condition: 26 UNIT 1  Electromagnetic Radiation and Atomic Spectra nh mvr = (1.12) 2π Solving Equation 1.12 for v gives: nh v= (1.13) 2π mr Substituting for v in Equation 1.11 and solving for r gives: n2 h2 2 h2  (1.14) r= = n  2  4πme 2  4πme  Here, n is positive integer (n = 1, 2, 3...) and is called quantum number. It is known that h, π, m, and e are constants, thus Equation 1.14 can be simplified to: r = n2 ao (1.15) According to Equation 1.15 the only permitted values of the radii of the electron path in the hydrogen atom are those proportional to the square of a whole number, n. The numerical values of ao is 0.53 Å. Thus, r = (0.53 Å)n2 For instance, for n = 1, r = 0.53 Å. This is the first Bohr radius. The larger the values of n, the further the electron from the nucleus. Figure 1.13: Bohr’s energy levels of a hydrogen atom UNIT 1 27  CHEMISTRY GRADE 11 Bohr showed that the energies that an electron in hydrogen atom can occupy are given by: R En = - H2 (1.16) n Where RH, the Rydberg constant for the hydrogen atom and has the value 2.18 x 10-18 J. Thus, Equation 1.15 can be written as: 2.18 × 10−18 J En = − n2 The negative sign in Equation 1.16 is an arbitrary convention, signifying that the energy of the electron in the atom is lower than the energy of a free electron, which is an electron that is infinitely far from the nucleus. The energy of a free electron is given a value of zero. As the electron gets closer to the nucleus (as n decreases), En becomes larger in absolute value, but also more negative. The most negative value, then, is reached when n = 1, which corresponds to the most stable energy state. We call this the ground state or ground level, which refers to the lowest energy state of an atom. The stability of the hydrogen electron diminishes for n = 2, 3.... Each of these levels is called an excited state, or excited level, which is higher in energy than the ground state. A hydrogen electron for which n is greater than 1 is said to be in an excited state. Using Bohr’s Equation 1.16, relating the energy (En ) and energy level (n) for an electron it is possible to calculate the energy of a single electron in a ground state or excited state, or the energy change when an electron moves between two energy levels. 28 UNIT 1  Electromagnetic Radiation and Atomic Spectra Example 1.6 Consider the n = 5 state of hydrogen atom. Using the Bohr model, calculate the radius of the electron orbit, the velocity and the energy of the electron. Solution: To determine the radius, Equation 1.15 is used: =r ( 0.53 Å ) × = 52 15.24=Å 1.524 nm Velocity of the electron is determined using Equation 1.13: nh v= 2πmr = ( 5 6.63 × 10−34 kg m 2 s −1 ) −31 2 × 3.14 × 9.11 × 10 kg × 1.524 × 10−9 m = 3.8 × 105 m / s Equation 1.16 is used to solve E5 2.18 × 10−18 J E5 = − 8.72 × 10−20 J = 5 2 Calculate the energies of the hydrogen electron in n = 1, n = 2 and n = 3 Solution: Using Equation 1.16: 2.18 × 10−18 J For n = 1, E1 = − − 2.18 × 10−18 J = 12 2.18 × 10−18 J n = 2, E2 = − − 5.54 × 10−19 J = 22 2.18 × 10−18 J 3 , E3 = n= − =− 2.42 × 10−19 J 32 Bohr’s model also quantitatively explained the line spectra of hydrogen atom. He proposed that the absorptions and emissions in line spectra correspond to the transfer of the electron from one orbit to another. Energy must be absorbed for the electron to move from one orbit to another one having a bigger radius. Whereas, energy is emitted when an electron moves from the higher orbital energy level, ni, to the lower orbital, nf, energy level. Thus, the change in energy, E, is the difference in the energy between the final and the initial state electron: UNIT 1 29  CHEMISTRY GRADE 11 ∆E = Ef − Ei (1.17) This equation is similar to:  1 1  ∆E =− RH  2 − 2   nf ni  (1.18) −18  1 1  = −2.18 × 10 J  2− 2  nf ni  Where ni and nf represent quantum numbers for initial and final states respectively. But, ∆E = hν, thus we have:  1 1  E hv = − 2.18 × 10−18 J  2 − 2  ∆= (1.19)  nf ni  Notice that when nf > ni, ∆E is positive, indicating that the system has absorbed energy. But, ni > nf, ∆E is negative and this corresponds to emission of energy. Example 1.7 Calculate the energy emitted when an electron moves from the n = 3 to the n = 2 energy level. Determine the wavelength of the emitted energy. Solution: Equation 1.19 is used, to determine the emitted energy:  1 1 ∆E = −2.18 × 10−18 J  2 − 2  =3.03 × 10−19 J 2 3  To obtain λ, Equation 1.19: hc ∆E = hv = λ m 6.63 × 10−34 J s × 3.0 × 108 hc s Thus, = = ∆E 3.03 × 10−19 J = 6.56 × 10−7 m = 656 nm 30 UNIT 1  Electromagnetic Radiation and Atomic Spectra Exercise 1.6 1. What is the wavelength of a photon (in nanometers) emitted during a transition from the n = 5 state to the n = 2 state in the hydrogen atom? 2. Calculate the frequency of the green line arising from the electron moving from n = 4 to n = 20 in the visible spectrum of the hydrogen atom using Bohr’s theory. Each spectral line in the emission spectrum corresponds to a particular transition in a hydrogen atom. When we study a large number of hydrogen atoms, we observe all possible transitions and hence the corresponding spectral lines. For instance, Figure 1.14 illustrates line spectra of a hydrogen atom when its electron moves from n = 4 to n = 1; n = 3 to n = 1 and n = 2 to n = 1. Figure 1.14: The emission spectrum of a hydrogen atom Each horizontal line represents an allowed energy level for the electron in a hydrogen atom. The energy levels are labeled with their principal quantum numbers. The emission spectrum of hydrogen includes a wide range of wavelengths from the infrared to the ultraviolet. Table 1.2 shows the series of transitions in the hydrogen spectrum; they are named after their discoverers. The Balmer series was particularly easy to study because some its lines fall in the visible range. UNIT 1 31  CHEMISTRY GRADE 11 Table 1.2: The various series in atomic hydrogen emission spectrum Series nf ni Spectrum region Lyman 1 2, 3, 4, … Ultraviolet Balmer 2 3, 4, 5, … Visible and ultraviolet Paschen 3 4, 5, 6,... Infrared Brackett 4 5, 6, 7,... Infrared For a larger orbit radius (i.e. a higher atomic energy level), the further the electron drops, the greater is the energy (higher v, shorter λ) of the emitted photon. Exercise 1.7 1. Calculate the energies of the states of the hydrogen atom with n = 2 and n = 3, and calculate the wavelength of the photon emitted by the atom when an electron makes a transition between these two states. 2. What is the wavelength of a photon emitted during a transition from the ni = 10 state to the nf = 2 state in the hydrogen atom?    The Bohr Model was an important step in the development of atomic theory. He introduced the idea of quantized energy states for the electron in a hydrogen atom. The model explains atoms and ions containing only one electron such as H, He+ and Li2+. However, the model has several limitations: It doesn't explain the atomic spectra of more complicated atoms and ions, even that of helium, the next simplest element. It doesn't explain about further splitting of spectral lines in the hydrogen spectra on application of a magnetic field. It considers electrons to have both known radius and orbit, which is impossible according to Heisenberg's uncertainly principle. 32 UNIT 1  Electromagnetic Radiation and Atomic Spectra       Bohr’s model assumed that an atom has only certain allowable energy levels in order to explain the observed line spectrum. However, his assumption had no basis in physical theory. In 1924, Louis de Broglie, the French physicist, proposed a surprising reason for fixed energy levels: if energy is particle-like, perhaps matter is wavelike. De Broglie reasoned that if electrons have wavelike motion and are restricted to orbits of fixed radii that would explain why they have only certain possible frequencies and energies. By combining Einstein’s equation for the quantity of energy equivalent to a given amount of mass (E = mc2) with Planck’s equation for the energy of a photon (E = hν = hc/λ), de Broglie derived an equation for the wavelength of any particle of mass, m, whether a planet, ball, or electron-moving at speed, v. h λ= (1.20) mv According to this Equation 1.20, matter and energy show both wave and particulate properties. The distinction between a particle and a wave is only meaningful in the macroscopic world, it disappears at the atomic level. This dual character of matter and energy is known as the wave-particle duality. Example 1.8 Calculate the de Broglie wavelength of an electron that has a velocity of 1.00 × 106 m/s. (electron mass, me = 9.11 × 10–31 kg; h = 6.63 × 10–34 J.s). m2 6.63 × 10−34 kg h s Solution: λ = = me v 1.00 × 106 m × 9.11 × 10−31 kg s = 0.73 × 10−9 m =0.728 nm What is the speed of an electron that has a de Broglie wavelength of 1.00 nm m2 Solution: 6.63 × 10−34 kg h s =v = me λ 1.00 × 10−9 m × 9.11 × 10−31 kg m m = 0.728 × 106 = 7.28 × 105 s s UNIT 1 33  CHEMISTRY GRADE 11 Exercise 1.8 1. What is the characteristic wavelength of an electron (in nm) that has a velocity of 5.97 ´ 10 6 m/s (m e = 9.11 ´ 10–31 kg)? 2. Calculate the wavelength (in nanometers) of an H atom (mass = 1.67 ´ 10-27 kg) moving at 7.00 ´ 10 2 cm/s. 3. Calculate the wavelength (in nm) of a photon emitted when a hydrogen atom undergoes a transition from n = 5 to n = 2. 1.6  At the end of this section, you will be able to: ) state Heisenberg's uncertainty principle ) describe the significance of electron probability distribution. ) explain the quantum numbers n, l, ml, ms ) write all possible sets of quantum numbers of electrons in an atom. ) describe the shapes of orbitals designated by s, p and d. Activity 1.13 Form a group and discuss the following questions, then present your responses to the rest of the class. 1. If particles have wavelike motion, why can’t we observe its motion in the macroscopic world? 2. If electrons possess particle nature it should be possible to locate electrons. How can an electron be located? 3. Is there any wave associated with a moving elephant? The Bohr model of an electron orbiting around the nucleus, looking like a planet around the Sun, doesn't explain properties of atoms. The planetary view of one charged particle orbiting another particle of opposite charge does not match some of the best-known laws of classical physics. 34 UNIT 1  The Quantum Mechanical Model of the Atom Scientists have developed quantum mechanics, which presents a different view of how electrons are arranged about the nucleus in the atom. This view depends on two central concepts: the wave behavior of matter and the uncertainty principle. The combination of these two ideas leads to a mathematical description of electronic structure.    Discovery of the wave properties of matter raised a new and very interesting question. If a subatomic particle can exhibit the properties of a wave, is it possible to say precisely where that particle is located. A wave, extends in space, so its location is not defined precisely. In 1827, Werner Heisenberg, the German physicist, proposed the Heisenberg uncertainty principle, stating that it is not possible to know with great certainty both an electron’s position and its momentum p (where p = mν) at the same time. Heisenberg was able to show that if ∆p is the uncertainty in the momentum and ∆x is the uncertainty of the position of the electron, then: h h ( ∆x )( ∆p ) ≥ or ( ∆x ) ∆ ( mv ) ≥ 4π 4π This equation shows that if we measure the momentum of a particle precisely (i.e., if ∆p is very small), then the position will be correspondingly less precise (i.e, ∆x will become larger). In simpler terms, if we know precisely where a particle is, we cannot know precisely where it has come from or where it is going. If we know precisely how a particle is moving, we cannot know precisely where it is. The Heisenberg uncertainty principle has no practical consequence in everyday objects in the macroscopic world. However, it has very significant consequences when applied to atomic and subatomic particles. UNIT 1 35  CHEMISTRY GRADE 11   The quantum mechanical description of the atom is based on the assumption that there are waves associated with both matter and radiations. The mathematical description of such wave provides information about the energy of electrons and their position. Erwin Schrödinger (1927), suggested that an electron or any other particle exhibiting wavelike properties can be described by a mathematical equation called a wave function (denoted Greek latter psi, ψ). The equation is very complex. However, it leads to series of solutions that describe the allowed energy states of electrons. The square of a wave function, ψ2, gives the probability of finding an electron in a certain region of a space. The results of a quantum mechanics indicate that the electron may be visualized as being in a rapid motion within a given region of space around the nucleus. Although we cannot determine the precise position of an electron, the probability of the electron being at a definite location can be calculated. The region in which an electron is most likely found is called an orbital. The electron may occupy anywhere within an orbital at any instant in time. So, an electron can be considered as a particle that can rapidly move from place to place, behaving like an “electron cloud” whose density varies within the orbital. In quantum mechanics, each electron in an atom is descried by four quantum numbers. Three of the quantum numbers specify the wave function that gives the probability of finding an electron at various place in spaces. The fourth one is used to describe the spin (magnetic property) of the electrons that occupy the orbitals. The four quantum numbers are: 1. The principal quantum number (n) describes the main energy level, or shell, an electron occupies. It may be any positive integer, n = 1, 2, 3, 4, etc. It describes the size and energy of the shell in which the orbital resides and it is analogous to the energy levels in Bohr’s model. 2. The angular momentum quantum number (ℓ) designates the shape of the atomic orbitals. Within a shell different sublevels or subshells are possible, each with a characteristic shape. It takes values from 0 to n-1. Orbitals of the same n, but 36 UNIT 1  The Quantum Mechanical Model of the Atom different ℓ are said to belong to different subshells. It is also called azimuthal quantum number or subsidiary quantum number. A given energy level, n, has n2 total number of orbitals. For example, in n = 4, ℓ has values of 0, 1, 2, and 3. Each value of ℓ corresponds to an orbital label and an orbital shape. The following letters are usually used to denote the values of ℓ: Value of ℓ Corresponding subshell 0 s 1 p 2 d 3 f Example 1.9 What do “4” and “d” represent in 4d? Solution: The number 4 gives the principal quantum number (n = 4) and the letter d gives the type of orbital (ℓ = 2). 3. The magnetic quantum number (mℓ) is also called the orbital-orientation quantum number. It has integral values between -ℓ and ℓ, including 0. The value of mℓ is related to the orientation of the orbital in space relative to the other orbitals in the atom. The number of possible mℓ values or orbitals for a given ℓ value is 2ℓ + 1. 4. The electron spin quantum number (ms) refers to the spin of an electron and the orientation of the magnetic field produced by this spin. For every set of n, ℓ, and mℓ values, ms can take the value +½ or -½. Each atomic orbital cannot accommodate more than two electrons, one with ms = +½ and another with ms = -½. UNIT 1 37  CHEMISTRY GRADE 11 Exercise 1.9 1. What are the values of n and ℓ for the following subshells? a. 1s c. 4s e. 4f b. 3p d. 3d 2. Write the subshell notations that correspond to a. n = 3 and ℓ = 0 c. n = 7 and ℓ = 0 b. n = 3 and ℓ = 1 d. n = 3 and ℓ = 2 Example 1.10 1. What are the allowed values of the quantum numbers through n = 2? Solution: The allowed quantum number values in n = 2 are: Electron capacity Electron capacity of n ℓ mℓ ms of subshell 4ℓ + 2 shell 2n2 1 0 (1s) 0 +½, -½ 2 2 2 0 (2s) 0 +½, -½ 2 8 2 1(2p) ±½ for each -1, 0, 1 value of mℓ 6 8 2. What values of the angular momentum quantum number (ℓ) and magnetic quantum number (mℓ) are allowed for a principal quantum number, n = 3? How many orbitals are allowed for n = 3? Solution: Determining ℓ values: For n = 3, ℓ = 0, 1, 2 Determining mℓ for each ℓ value: For ℓ = 0, mℓ = 0 For ℓ = 1, mℓ = –1, 0, +1 For ℓ = 2, mℓ = –2, –1, 0, +1, +2 Number of orbitals in n = 3 is n2 = 32 = 9 orbitals These orbitals are: 3s: 1 orbital 3p: 3 orbitals 3d: 5 orbitals Total = 9 orbitals 38 UNIT 1  The Quantum Mechanical Model of the Atom Exercise 1.10 1. What are the allowed values of the quantum numbers through n = 4? 2. Write an orbital designation corresponding to the quantum numbers: a. n = 4, ℓ = 2, m ℓ = 0 c. n = 5, ℓ = 1, m ℓ = -1 b. n = 3, ℓ = 1, m ℓ = 1 d. n = 4, l = 3, mℓ = -3 3. What values ℓ and m ℓ are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3? 4. Can an orbital have the quantum number n = 2, ℓ = 2, m ℓ = 2 5. For an orbital with n = 3 and m ℓ = 1, what is (are) the possible value(s) ℓ? 6. Which of the following orbitals do not exist: 1p, 2s, 2d, 3p, 3d, and 3f?     The three-dimensional aspects of the orientation of the atomic orbitals are usually represented by drawing a boundary surface diagram that encloses the highest probability (about 90%) of the total electron density in an orbital. Figure 1.15, shows that the region of the greatest probability of finding an s electron is in a spherically symmetrical space whose origin is the atomic nucleus. The 1s orbital is the smallest, the 2s orbital is larger than 1s and so on. Regardless of their principal quantum numbers, all s orbitals are spherical. Figure 1. 15: The three S-orbitals UNIT 1 39  CHEMISTRY GRADE 11 The p orbitals of any principal quantum number are arranged along three mutually perpendicular axes, x, y, and z, so that the region of the highest electron density are in dumb bell-shaped boundary surface (Figure 1.16). The three p orbitals are designated as px, py and pz. Figure 1. 16: The three p - orbitals Two lobes of each p orbital lie along a line with the nucleus at their center. For instance, the three 2p orbitals are classified as 2px, 2py and 2pz Higher principal energy levels (n > 3) have five d orbitals in addition to, one s orbital, three 3p orbitals. The special orientations of d orbitals are much more complex in shape than p orbitals. Figure 1.17 shows the different distribution of the five d atomic orbitals. Figure 1.17: The five d orbitals 40 UNIT 1  Electronic Configurations and Orbital Diagrams The principal energy level whose values are n ≥ 4 have additional fourth sublevels. For instance, in n = 4, there are a 4s subshell with one orbital, a 4p subshell with three orbitals, a 4d subshell with five orbitals, and a 4f subshell with seven orbitals. The f orbitals have more complex shapes than the d orbitals. 1.7     At the end of this section, you will be able to: ) explain the Aufbau Principle ) explain Pauli's Exclusion Principle ) explain Hund’s Rule ) write ground state electron configurations of multi-electron atoms. Activity 1.14 Form a group and discuss the given questions. Then, share your ideas with the rest of the class. 1. Why dosen't an atomic orbital hold more than two electrons? 2. Why do you think that electrons in an atom could not occupy the spaces between the main energy levels? Several questions arise when you look carefully at the electron configuration of an atom. To answer these questions, it is necessary to know the basic principles governing the distribution of electrons among atomic orbitals. The electron configuration for any atom is governed by the following three principles: 1. Aufbau (building up) Principle: This is a scheme used to reproduce the electron configurations of the ground state of atoms by successively filling with electrons in a specific order (the building up order). In general, electrons occupy the lowest-energy orbital available before entering the higher energy orbital. Accordingly, the ground state electron configurations of atoms are obtained by filling the subshell in the following order; 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, and so on, as indicated in Figure 1.18. UNIT 1 41  CHEMISTRY GRADE 11 Figure 1.18: The order of filling orbitals 2. Hund’s Principle: Equal energy orbitals (degenerate orbitals) are each occupied by a single electron before the second electrons of opposite spin enters the orbital. In other words, each of the three 2p orbitals (2px, 2py and 2pz) will hold a single electron before any of them receives a second electron. 3. Pauli’s Exclusion Principle: No two electrons can have the same four quantum numbers. That means, they must differ in at least one of the four quantum numbers.       The electronic configuration of an atom describes the distribution of the electrons among atomic orbitals in the atom. Two general methods are used to denote electron configurations. The subshell (sublevel) notation uses numbers to designate the principal energy levels and the letters s, p, d and f to identify the sublevels. A superscript number following a letter indicates the number of electrons (e) in the designated subshell. The designation can be written as nle. For instance, the ground state electron configuration of H (Z =1) is 1s1. 42 UNIT 1  Electronic Configurations and Orbital Diagrams Activity 1.15 Form a group and discuss the reason why the notation nle does not include electron spin quantum numbers. (Where n, l and e are principal quantum number, azimuthal quantum number and number of electrons, respectively). The other way to present this information is through an orbital diagram, which consists of a rectangular box (or circle, or just a line) for each orbital available in a given energy level, with an arrow indicating the electron’s presence and its direction of spin. Traditionally, ↑ represents, ms = +½ whereas ↓ designates ms = -½. The orbital diagrams for the first ten elements are: H (Z=1) He (Z=2) Li (Z=3) 1s1 1s2 1s2 2s1 2p Be (Z=4) B (Z=5) 1s2 2s2 2p 1s2 2s2 2p1 C (Z=6) N (Z=7) 1s2 2s2 2p2 1s2 2s2 2p3 O (Z=8) F (Z=9) 1s2 2s2 2p4 1s2 2s2 2p5 Ne (Z=10) 1s2 2s2 2p6 The three orbitals in the 2p sublevel are degenerate (have equal energy) and have the same n and ℓ values), thus, the fifth electron of boron can go into any one of the 2p orbitals. Since a p sublevel has ℓ = 1, the mℓ (orientation) values can be –1, 0, or +1. For convenience, the boxes are labeled from left to right, –1, 0, +1, and assume it enters the mℓ = –1 orbital: n = 2, ℓ = 1, ml = –1, and ms = +½. The three 2p orbitals have equal energy, but they can be along the x, y, or z axes. UNIT 1 43  CHEMISTRY GRADE 11 Activity 1.16 Discuss the following questions in groups, and share your ideas with the rest of the class. 1. What does each box in an orbital diagram represent? 2. What are the similarities and differences between a 1s and a 2s orbital? 3. What is the difference between a 2p x and a 2p y orbital? 4. What is the meaning of the symbol 4d 6? Exercise 1.11 1. Write the expected electron configuration for these atoms: a. Na (Z= 11) d. V (Z = 23) b. Al (Z = 13) e. Mn (Z =25) c. P (Z =15) f. Fe (Z = 26) 2. Indicate the total number of: a. p electrons in N (Z = 7) c. 3d electrons in S (Z = 16) b. s electrons in Si (Z = 14) 3. Indicate which of the following sets of quantum numbers in an atom are unacceptable and explain why: a. (1, 0, ½, ½) d. (2, 2, 1, +½) b. (3, 0, 0, +½) ` e. (4, 3, -2, +½) c. (3, 2, 1, 1) For Li, the first 2 electrons occupy the 1s orbital, and the third electron must occupy the first orbital with n = 2, the 2s orbital. Thus, the electron configuration for Li is 1s22s1. To avoid writing the inner-level electrons, this configuration is often abbreviated as [He] 2s1, where [He] represents the electron configuration of He, 1s2. Similarly, Mg (Z =12), has the configuration 1s22s22p63s2, or [Ne] 3s2. Then the next six elements, aluminium to argon, have configurations obtained by filling the 3p orbitals one electron at a time. Table 1.3 shows the ground state electron configurations of some atoms. 44 UNIT 1  Electronic Configurations and Orbital Diagrams Table 1.3: Ground state electron configurations of some atoms Atom Atomic number (Z) Electron configuration H 1 1s1 He 2 1s2 Be 4 1s22s2 or [He]2s2 O 8 1s22s22p4 or [He]2s22p4 Ne 10 1s22s22p6 or [He]2s22p6 Cl 17 1s22s22p63s23p5 or [Ne] 3s23p5 Ar 18 1s22s22p63s23p6 or [Ne] 3s23p6 Ca 20 1s22s22p63s23p64s2 or [Ar] 4s2 Sc 21 1s22s22p63s23p64s2 3d1or [Ar] 4s23d1 Cr 24 1s22s22p63s23p64s1 3d5 or [Ar] 4s13d5 Notice that in the electron configuration of an atom, the electrons in the outermost principal quantum level of an atom are called valence electrons. For example, the valence electrons of the Be atom, are the 2s electrons. For the Ne atom, the valence electron is the electrons are in the 2s and 2p electrons. Valence electrons are the most important electrons to chemists since they are involving in bonding, as you will learn in the next chapter. The inner electrons are known as core electrons. Examine the electron configurations of chromium and copper. The expected configurations, those based on the Aufbau principle, are not the ones observed through the emission spectra and magnetic properties of these elements. Expected Observed Cr (Z = 24 ) [Ar]4s23d4 [Ar]4s13d5 Cu (Z = 29 ) [Ar]4s23d9 [Ar]4s13d10 The reason for these exceptions to the Aufbau principle are not completely understood, but it seems that the half-filled 3d subshell of chromium (3d5) and the fully filled 3d subshell of copper (3d10) lends a special stability to the electron configurations. Apparently, having a half-filled 4s subshell and a half-filled 3d subshell gives a lower energy state for a Cr atom than having a filled 4s subshell. UNIT 1 45  CHEMISTRY GRADE 11 Because there is little difference between the 4s and 3d orbital energies, expected and observed electron configurations are quite close in energy. At higher principal quantum numbers, the energy difference between certain subshell is even smaller than that between the 3d and 4s subshells. As a result, there are still more exceptions to the Aufabu principle among the heavier transition elements. Activity 1.17 Form a group and complete the following table for the first-row transition metals on your note book: Element Atomic Ground state electron Condensed electron number configuration configuration Sc 21 Ti 22 V 23 Cr 24 Mn 25 Fe 26 Co 27 Ni 28 Cu 29 Zn 30 46 UNIT 1  Electronic Configurations and the Periodic Table of the Elements 1.8        At the end of this section, you will be able to: ) correlate the electron configurations of elements with their positions in the periodic table ) give a reasonable explanation for the shape of the periodic table ) classify elements as representative, transition and inner-transition elements ) explain the general trends in atomic radius, ionization energy, electron affinity, electronegativity, and metallic character of elements within a period and group of the periodic table ) write the advantages of the periodic classification of elements.      Activity 1.18 In your Grade 9 chemistry lessons, you have learned about the development of the modern periodic table. Form a group and discuss the following questions, then present your responses to the whole class. 1. How is today’s periodic table different from the one that Mendeleev published? 2. What is a “period?” What does it represent? 3. What is a “group?” What does it represent? 4. How does the electron configuration of an element give information about the period and group it is in? 5. Why are some elements placed out of the main body of the periodic table? UNIT 1 47  CHEMISTRY GRADE 11 Elements are arranged in the modern periodic table in a specific pattern that helps to predict their properties and to show their similarities and differences. Periodic relationships can be summarized by the general statement called the periodic law. In its modern form, the periodic law states that certain sets of physical and chemical properties repeat at regular intervals (periodically) when the elements are arranged according to increasing atomic number.     Activity 1.19 Answer the following questions in groups and present your responses to the whole class. 1. What is the importance of classifying elements in groups and periods? 2. What could be the basis of classifying elements? 3. Why are there two different numbering systems for groups? 4. Which sublevel is being filled in period 1? The periodic classification follows as a logical consequence of the electronic configuration of atoms. Elements are grouped according to their outer-shell electron configurations, which account for their similar chemical behavior. The period indicates the value of n for the outermost or valence shell. There are 18 groups and 7 periods in the modern periodic table. A metalloid is an element that has properties that are between those of metals and non-metals, for example boron, germanium and silicon. 48 UNIT 1  Electronic Configurations and the Periodic Table of the Elements Project 1.1 Form a group of five and do the following activities and submit a report both in word and PDF formats. The elements can be divided into representative, transition, and inner transition elements based on the type of subshell being filled. Discuss these different classes of elements by searching information from the internet or any other sources. Also, identify each block in the blank periodic table.   Activity 1.20 Discuss the following questions in groups and, present your responses to the whole class. 1. How does size of elements vary across a period? 2. Why are elements of Groups 1 (IA) and 2 (IIA) called metals, while those of Group 17 (VIIA) non-metals? 3. Why are elements of Group 18 (VIIIA) least reactive? The electron configurations of the atoms display a periodic variation with increasing atomic number (nuclear charge). As a result, the elements show periodic variations of physical and chemical behavior. In this section you will examine some periodic atomic properties like atomic radii, ionization energies, electron affinities, electronegativity, and metallic character. UNIT 1 49  CHEMISTRY GRADE 11 Atomic Size (Atomic Radii) Activity 1.21 Answer the following questions in groups, then present your responses to the whole class. 1. Why don't the sizes of atoms increase uniformly with increasing atomic number? 2. In which location in the periodic table would you expect to find the elements having the largest atoms? Explain. 3. How do the sizes of atoms change: a. from left to right in a row in the periodic table? b. from top to bottom in a group in the periodic table? Explain. Exact size of an isolated atom cannot be measured because the electron cloud surrounding the atom does not have a sharp boundary. However, an estimate of the atomic size can be made by measuring the distance between the nuclei of two adjacent atoms. This property is called the atomic radius. One of the most common methods to determine the atomic radius is to assume that atoms are spheres touching each other when they are bonded together (Figure 1.19). Figure 1.19: Atomic radius for a diatomic molecule. d r= 2 Atomic size varies within both a group and a period of the main-group elements. These variations in atomic size are the result of two opposing influences: changes in the principal quantum number (n) and changes in the effective nuclear charge ( Zeff ) the nuclear charge an electron actually experiences. The net effect of these depends on the shielding of the increasing nuclear charge by the inner electrons. 50 UNIT 1  Electronic Configurations and the Periodic Table of the Elements Activity 1.22 Discuss the following questions in groups, and present your responses to the whole class. Compare the effect of Z eff and n on atomic size. Which one dominates? a. down a group? b. across a period? Explain the reason. Atomic radius generally increases in a group from top to bottom but it decreases in a period from left to right. These trends hold well for the main-group elements but not as for the transition elements (electrons enter an inner electron shell, not the valence shell). As we move from left to right, size shrinks through the first two or three transition elements because of the increasing nuclear charge. But, from then on, the size remains relatively constant because shielding by the inner d electrons cancels out the usual increase in Zeff. For instance, vanadium (V; Z =23), the third Period 4 transition metal, has the same atomic radius as zinc (Zn; Z = 30), the last Period 4 transition metal. Exercise 1.12 1. What are the factors affecting the size of an atom of an element? Explain your answer using examples. 2. Using the periodic table, rank each set of main-group elements in order of decreasing atomic size: a. Ca, Mg, Sr b. K, Ga, Ca c. Br, Rb, Kr d. Sr, Ca, Rb 3. Why is the difference in atomic radius between the elements Z = 11 (Na; 186 pm) and Z = 12 (Mg; 160 pm) so large, whereas between Z = 24 (Cr; 125 pm) and Z = 25 (Mn; 124 pm) the difference is very small. (Note that "pm" is picometers.). UNIT 1 51  CHEMISTRY GRADE 11 Ionization Energy (IE) Activity 1.23 Form a group and discuss the following, then present your responses to the whole class. 1. Write the electron configuration of the following pairs of atoms of an element and compare their stability. Explain your answer. a. oxygen and nitrogen b. magnesium and aluminum c. fluorine and neon 2. From which atom in each pair (question number 1) will it be easier to remove an electron from the outer most shell? Explain your answer based on their electron configurations. The ionization energy (IE) is the energy (in kJ) required for the complete removal of 1 mol of electrons from 1 mol of gaseous atoms or ions. Multi-electron atoms can lose more than one electron, so the ionization energies required to remove each electron are numbered in sequence from the ground-state atom. For instance, boron atom has five electrons, two in an inner core (1s2) and three valence electrons (2s22p1). The five ionization steps and their ionization energies, IE1 through IE5, are: B (g) → B+ (g) + e– IE1 = 801 kJ mol–1 B+ (g) → B2+ (g) + e– IE2 = 2427 kJ mol–1 B2+ (g) → B3+ (g) + e– IE3 = 3660 kJ mol–1 B3+ (g) → B4+ (g) + e– IE4 = 25,025 kJ mol–1 B4+ (g) → B5+ (g) + e– IE5 = 32,822 kJ mol–1 The first ionization energy ( IE1 ) of an atom is the minimum energy needed to remove the first electron from the outermost shell of a gaseous atom to produce a gaseous ion. Atom (g) → Ion+(g) + e– ∆E = IE1 > 0 52 UNIT 1  Electronic Configurations and the Periodic Table of the Elements The second ionization energy (IE2) removes the second electron. Since the electron is being pulled away from a positively charged ion, IE2 is always larger than IE1: Ion+ (g) → Ion2+ (g) + e– ∆E = IE2 (always > IE1) The first ionization energy is a key factor in an element’s chemical reactivity because, atoms with a low IE1 tend to form cations during reactions, whereas those with a high IE1, (except the noble gases) often form anions. Ionization energies display a periodic variation when plotted against atomic number, as Figure 1.20 shows. Within any period, values tend to increase with atomic number. Thus, the lowest values in a period are found for the Group IA elements (alkali metals). It is characteristic of reactive metals such as these to lose electrons easily. The largest ionization energies in any period occur for the noble-gas elements. This general trend increasing ionization energy with atomic number in a given period is due to the fact that as we move across a period, Zeff generally increases so atomic radii become smaller. As a result, the attraction between the nucleus and the outer electrons increases, so an electron becomes more difficult to remove. Figure 1.20: Ionization energy versus atomic number Small deviations from this general trend occur. A Group IIIA element (ns2np1) has a smaller ionization energy than the preceding Group IIA element (ns2). UNIT 1 53  CHEMISTRY GRADE 11 Apparently, the np electron of the Group IIIA element is more easily removed than one of the ns electrons of the preceding Group IIA element. Also note that a Group VIA element (ns2np4) has a smaller ionization energy than the preceding Group VA element. As a result of electron repulsion, it is easier to remove an electron from the doubly occupied np orbital of the Group VIA element than from a singly occupied orbital of the preceding Group VA element. Group Assignment 1.1 Form a group and discuss on the following questions. Write your reflections and submit it to your teacher. 1. The first ionization energy in general increases across a period, however, large drops occur when a new period begins. Why? 2. Explain the irregularities in the trends across periods: a. Boron has a smaller first ionization energy than beryllium. b. The first ionization energy of nitrogen is higher than oxygen. Exercise 1.13 1. Choose the element with the higher ionization energy from each pair: a. As or Bi b. As or Br c. Al or In d. K or Ge 2. The first and second ionization energies of potassium, K, are 419 kJ mol–1 and 3052 kJ mol–1 and those of calcium, Ca, are 590 kJ mol–1 and 1145 kJ mol –1, respectively. Compare their values and comment on the differences. 3. Based on their positions in the periodic table, predict which atom of the following pairs will have the larger first ionization energy: a. Ga or Ge b. Br or Sb c. K or Cr d. Mg or Sr e. O or Ne 4. Name the period 3 element with the following ionization energies (in kJ/mol), and write its electron configuration: IE1 IE2 IE3 IE4 IE5 IE6 1012 1903 2910 4956 6278 22,330 54 UNIT 1  Electronic Configurations and the Periodic Table of the Elements Electron Affinity (EA) The electron affinity is the energy change for the process of adding an electron to a neutral atom in the gaseous state to form a negative ion: Atom (g) + e−  → Ion− (g) ∆ = EA1 An electron approaching a neutral atom experiences an attraction for the positively charged nucleus. Repulsion of the incoming electron by electrons already present in the atom tends to offset this attraction. Still, in many cases the incoming electron is absorbed by the atom and energy is evolved in the process. Thus, first electron affinity (EA1) is usually negative. The second electron affinity (EA2), however, is always positive because energy must be absorbed to overcome electrostatic repulsions and add another electron to a negative ion. For example, the EA1 and EA2 of an oxygen atom can be written as: O(g) + e– → O– (g) EA1 = –141 kJ mol–1 O–(g) + 1e– → O2–(g) EA2 = +744 kJ mol–1 Figure 1.21 gives the electron affinities of the main-group elements. Negative values indicate that energy is released when the anion forms. Positive values, which occur in Group VIIIA (18), indicate that energy is absorbed to form the anion; in fact, these anions are unstable and the values are estimated. Figure 1. 21: Electron affinities of the main-group elements (in kJ/mol) UNIT 1 55  CHEMISTRY GRADE 11 Activity 1.24 Discuss the following questions in groups, then present your answers to the whole class. 1. Which group in the periodic table has elements with high (endothermic) IE1 and very negative (exothermic) first electron affinities (EA1)? Give the charge on the ions these atoms form. 2. Silicon has an electron affinity of –134 kJ/mol. The electron affinity of phosphorus is –72 kJ/mol. Give a reason for this difference. 3. Why are the electron affinities of the alkaline earth metals either negative or small positive values? Electron affinities, EA, have a periodic variation, just as atomic radii and ionization energies do. Broadly speaking, the general trend is toward more negative electron affinities from left to right in any period. What could be the reason the IIA element and the VA element tend to have smaller electron affinities (Figure 1.21) than the preceding element? Note that the Group VIA and Group VIIA elements have the largest negative electron affinities of any of the main-group elements. The trend in going across a period is rather clear but there is no simple trend in going down a column of elements. In most cases, the added electron goes into an energy sublevel that is already partly filled. For Group IIA and VIIIA atoms, however, the added electron would be required to enter a significantly higher energy level, the np level for the Group IIA atoms and the s level for the next principal level for the Group VIIIA atoms. In these cases, a stable anion does not form. Electronegativity Electronegativity indicates the extent of attraction by which the electrons of the bond pair are attracted by an atom linked by this bond. Linus Pauling, an American scientist, in 1922 assigned arbitrarily a value of 4.0 to fluorine, the element considered to have the greatest ability to attract electrons. This is known as the Pauling scale. Approximate values for the electronegativity of a few elements are given in Table 1.4. 56 UNIT 1  Electronic Configurations and the Periodic Table of the Elements Table 1.4: (a) Electronegativity values (on Pauling scale) across the periods Atom (Period II) Li Be B C N O F Electronegativity 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Atom (Period III) Na Mg Al Si P S Cl Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Table 1.4: (b) Electronegativity values (on Pauling scale) down a group Atom Atom Electronegativity Electronegativity (Group IA) (Group VIIA) value value Li 1.0 F 4.0 Na 0.9 Cl 3.0 K 0.8 Br 2.8 Rb 0.8 I 2.5 Cs 0.7 At 2.2 Electronegativity generally increases across a period from left to right (say from lithium to fluorine) and decreases down a group (say from fluorine to astatine) in the periodic table. The attraction between the valence electrons and the nucleus increases as the atomic radius decreases in a period. The electronegativity also increases. On the same account electronegativity values decrease with the increase in atomic radii down a group. The trend is similar to that of ionization energy. Activity 1.25 Discuss the following questions in groups, and present your answers to the whole class. 1. What is the electronegativity of an atom? How is it different from electron affinity? 2. In a given family of the periodic table, what is the general relationship between electronegativity and size? UNIT 1 57  CHEMISTRY GRADE 11 Metallic Character Activity 1.26 Form a group and summarize the trend in metallic character in the periodic table. Is it the same as the trend observed for atomic size and ionization energy? Present your responses to the whole the class. Metallic character refers to the chemical properties associated with elements classified as metals. These properties arise from the elements, ability to lose electrons. Metallic character decreases as you move across a period in the periodic table from left to right. This occurs as atoms more readily accept electrons to fill a valence shell than lose them to remove the unfilled shell. Metallic character increases as you move down an element group in the periodic table. This is because electrons become easier to lose as the atomic radius increases, where there is less attraction between the nucleus and the valence electrons because of the increased distance between them.         Activity 1.27 Form a group and discuss the following questions, then share your answers with the whole class. 1. In which region of the periodic table are metals located? 2. In which region are the elements with general electronic configuration of ns2p5 located? Give the group number. 3. Write the group number and period of an element with atomic number 26. 4. Is an element with the electronic configuration 1s2 2s2 2p 6 3s1 a metal or a non- metal? Will it form a cation or an anion readily? Explain your answer. 58 UNIT 1  Electronic Configurations and the Periodic Table of the Elements Some of the advantages of periodic classification of elements are: 1. The classification of elements is based on the atomic number, which is a fundamental property of an element. 2. Isotopes are averaged in one place, as the classification is on the basis of atomic number (For example, chlorine (atomic number 17 ) is given the atomic mass 35.5.) 3. It explains the periodicity of the properties of the elements and relates them to their electronic configurations. 4. The position of the elements that were misfits on the basis of mass number (anomalous pairs like argon and potassium) could be justified on the basis of atomic number. 5. The lanthanides and actinides are placed separately at the bottom of the periodic table. 6. The table is a simple, systematic and an easy way of remembering the properties of various elements, as it is based on the electronic configuration. UNIT 1 59  CHEMISTRY GRADE 11 UNIT SUMMARY ~ Modern chemistry began with eighteenth century discoveries leading to the formulation of two basic laws of chemical combination: the law of conservation of mass and the law of constant composition (definite proportions). Dalton proposed another law of chemical combination, the law of multiple proportions. ~ The first clues to the structure of atoms came through the discovery and characterization of cathode rays (electrons). Key experiments were those that established the mass-to-charge ratio and then the charge on an electron. ~ The principal types of radiation emitted by radioactive substances are alpha particles, beta particles, and gamma rays. ~ Studies on the scattering of particles by thin metal foils (Rutherford’s atomic model) led to the concept of the nuclear atom − a tiny, but massive, positively charged nucleus surrounded by lightweight, negatively charged electrons. ~ Electromagnetic radiation is characterized by its wavelength (λ), frequency (ν), and speed (c = 3.0 × 108 m/s), which are related by the formula: c = λν ~ Electromagnetic radiation can be viewed as a stream of “particles” called photons, each with energy hν, where h is Planck’s constant (6.63 × 1034 J.s). ~ When light strikes a metal surface, electrons are emitted. ~ Analysis of the kinetic energy and numbers of the emitted electrons led Einstein to suggest that electromagnetic radiation can be viewed as a stream of photons. ~ The photoelectric effect is the emission of an electron from the surface of a metal, caused by electromagnetic radiation of a certain minimum energy; the resulting current increases with increasing intensity of radiation. ~ Bohr’s theory requires the electron in a hydrogen atom to be in one of a discrete set of energy levels. The fall of an electron from a higher to a lower energy level releases a discrete amount of energy as a photon of light with a characteristic frequency. ~ Bohr’s theory accounts for the observed atomic spectrum of hydrogen atom. 60 UNIT 1  Electronic Configurations and the Periodic Table of the Elements ~ An emission spectrum is the spectrum associated with the emission